Ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable in English | class 8 chapter 2 maths exercise 2.1, 2.2, 2.3

Ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable. From class 8 chapter 2 maths in english. Here in ncert solutions for maths chapter 2 class 8 we given all questions solutions in detail. NCERT Solution of class 8 subject math chapter 2 Linear Equations in One Variable get all questions answers, notes, pdf like format. In this ncert maths class 8 chapter 2 students will get to know the concept of Pair of Linear Equations in Two Variables. chapter 2 maths class 8 explains how to represent a situation algebraically and graphically. Students will explore the methods of solving ncert maths class 8 chapter 2 the pair of the linear equation through Graphical Method. This chapter describes the Algebraic Method, Elimination Method, Cross-Multiplication Method, Substitution Method respectively. Students must practice this chapter to master the method of solving the linear equations. The exercises present in the chapter should be dealt with utmost sincerity if one wants to score well in the examinations.

Students can practice the questions based on the same through ncert solutions for class 8 maths chapter 2 exercise 2.1, ncert solutions for class 8 maths chapter 2 exercise 2.2, ncert solutions for class 8 maths chapter 2 exercise 2.3 and many more. These examples are in the form of word problems.
The next topic is Solving equations having the variable on both sides. In this section, students will learn how to solve such equations which have expressions with the variable on both sides.

Here we solve ncert solutions for ncert maths class 8 chapter 2 solution From class 8 maths chapter 2 pdf concepts all solutions with easy method with expert solutions. chapter 2 maths class 8 help students in their study, home work and preparing for exam. Soon we provide CBSE chapter 2 maths class 8 solutions English Medium. NCERT solutions for class 8 maths chapter 2 pdf Linear Equations in One Variable PDF here. You can download ncert english book class 8 from official NCERT website or Click HERE.

ncert solutions for class 8 maths chapter 2
Linear Equations in One Variable

ncert solutions for class 8 maths chapter 2 Linear Equations in One Variable in English

NCERT solutions for class 8 maths chapter 2 pdf – Questions Answer Click Here

ncert solutions for class 8 maths chapter 2 EXERCISE 2.1

Solve the following equations.

1. 𝑥 – 2 = 7 2. 𝑦 + 3 = 10 3. 6 = 𝑧 + 2
4. \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\) 5. 6𝑥 = 12 6. \(\displaystyle \frac{t}{5}=10\)
7. \(\displaystyle \frac{{2x}}{3}=18\) 8. \(\displaystyle 1.6=\frac{y}{{1.5}}\) 9. 7𝑥 – 9 = 16
10. 14𝑦 – 8 = 13 11. 17 + 6𝑝 = 9 12. \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)

Solutions:

1. 𝑥 – 2 = 7
Sol.: 𝑥 – 2 = 7
𝑥 = 7 + 2 (by transposing)
𝑥 = 9 answer

2. 𝑦 + 3 = 10
Sol.:  𝑦 + 3 = 10
𝑦 = 10 – 3 (by transposing)
𝑦 = 7 answer

3. 6 = 𝑧 + 2
Sol.: 6 = 𝑧 + 2
or 𝑧 + 2 = 6
𝑧 = 6 – 2 (by transposing)
𝑧 = 4 answer

4. \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\)
Sol.: \(\displaystyle \frac{3}{7}+x=\frac{{17}}{7}\)
\(\displaystyle x=\frac{{17}}{7}-\frac{3}{7}\) (by transposing)
\(\displaystyle x=\frac{{17-3}}{7}\) (by L.C.M)
\(\displaystyle x=\frac{{14}}{7}\)

𝑥 = 2 answer

6. 6𝑥 = 12
Sol.: 6𝑥 = 12
\(\displaystyle x=\frac{{12}}{6}\)
𝑥 = 2 answer

7. \(\displaystyle \frac{{2x}}{3}=18\)
Sol.: \(\displaystyle \frac{{2x}}{3}=18\)
\(\displaystyle \frac{{2x}}{3}=\frac{{18}}{1}\)
2𝑥 × 1 = 18 × 3 (by cross multiplication)
2𝑥 = 54
\(\displaystyle \frac{{54}}{2}\)𝑥 = 27 answer

8. \(\displaystyle 1.6=\frac{y}{{1.5}}\)
Sol.: \(\displaystyle 1.6=\frac{y}{{1.5}}\)\(\displaystyle \frac{{1.6}}{1}=\frac{y}{{1.5}}\)𝑦 = 1.5 × 1.6
𝑦 = 2.4 answer

9. 7𝑥 – 9 = 16
Sol.: 7𝑥 – 9 = 16
7𝑥 = 16 + 9
7𝑥 = 25
\(\displaystyle \frac{{25}}{7}\) answer

10. 14𝑦 – 8 = 13
Sol.: 14𝑦 – 8 = 13
14𝑦 = 13 + 8
14𝑦 = 21
𝑦 = \(\displaystyle \frac{{21}}{{14}}\)
𝑦 = \(\displaystyle \frac{3}{2}\) answer

11. 17 + 6𝑝 = 9
Sol.: 17 + 6𝑝 = 9
6𝑝 = 9 – 17 (by transposing)
6𝑝 = – 8
\(\displaystyle \frac{{-8}}{6}\)
\(\displaystyle \frac{{-4}}{3}\) answer

12. \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)
Sol.: \(\displaystyle \frac{x}{3}+1=\frac{7}{{15}}\)
\(\displaystyle \frac{x}{3}=\frac{7}{{15}}-\frac{1}{1}\) (by transposing)
\(\displaystyle \frac{x}{3}=\frac{{7-1}}{{15}}\) (by L.C.M.)
\(\displaystyle \frac{x}{3}=\frac{6}{{15}}\)
15 × 𝑥 = 6 × 3 (by cross multiplication)
15𝑥 = 18
\(\displaystyle x=\frac{{18}}{{15}}\)
\(\displaystyle x=\frac{{6}}{{5}}\) answer

Ncert Solutions for class 8 maths chapter 2 EXERCISE 2.2

1. If you subtract \(\displaystyle \frac{1}{2}\) from a number and multiply the result by \(\displaystyle \frac{1}{2}\), you get \(\displaystyle \frac{1}{8}\). What is the number?
Sol.: Let’s the number is = 𝑥
subtract \(\displaystyle \frac{1}{2}\) from x = \(\displaystyle x-\frac{1}{2}\)
Result is \(\displaystyle \frac{{2x-1}}{2}\)
Result multiply by \(\displaystyle \frac{1}{2}\)
According to the question –
\(\displaystyle \begin{array}{l}\left( {\frac{{2x-1}}{2}} \right)\times \frac{1}{2}=\frac{1}{8}\\\frac{{2x-1}}{4}=\frac{1}{8}\end{array}\)
\(\displaystyle 8\times (2x-1)=4\times 1\) (by cross multifaction)
16𝑥 – 8 = 4 
16𝑥 = 4 + 8 (by transposing)
16𝑥 = 12
\(\displaystyle \begin{array}{l}x=\frac{{12}}{{16}}\\x=\frac{3}{4}\end{array}\)
so that number is \(\displaystyle \frac{3}{4}\) answer.

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Sol.: Let’s the breadth of the swimming pool is = 𝑥 
so Its length is = 2𝑥 + 2
Perimeter of a rectangle is = 154 m
2 (length + breadth) = 154
put the value –
2 (2𝑥 + 2 + 𝑥) = 154
2 (3𝑥 +2) = 154
6𝑥 + 4 = 154
6𝑥 = 154 – 4
6𝑥 = 150
\(\displaystyle x=\frac{{150}}{6}\)
𝑥 = 25
so the breadth of the pool is 𝑥 = 25 m
and the length of the pool is 2𝑥 + 2 
2 × 25 + 2 = 52 m

3. The base of an isosceles triangle is \(\displaystyle \frac{4}{3}\) cm. The perimeter of the triangle is \(\displaystyle 4\frac{2}{{15}}\) cm. What is the length of either of the remaining equal sides?
Sol.: Let’s the equal side is = 𝑥
Base of isosceles triangle is = \(\displaystyle \frac{4}{3}\) m
Perimeter of the triangle = \(\displaystyle 4\frac{2}{{15}}\)
Sum of all three sides = \(\displaystyle 4\frac{2}{{15}}\)
\(\displaystyle \begin{array}{l}x+x+\frac{4}{3}=4\frac{2}{{15}}\\2x+\frac{4}{3}=\frac{{62}}{{15}}\end{array}\)
\(\displaystyle 2x=\frac{{62}}{{15}}-\frac{4}{3}\) (by transposing)
\(\displaystyle \begin{array}{l}2x=\frac{{62-20}}{{15}}\\2x=\frac{{42}}{{15}}\\2x=\frac{{14}}{5}\\x=\frac{{14}}{5}\times \frac{1}{2}\\x=\frac{7}{5}\end{array}\)
The length of each of the remaining equal sides x = \(\displaystyle \frac{7}{5}\) cm
or \(\displaystyle 1\frac{2}{5}\) cm

4. Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Sol.: Let’s the first number is = 𝑥
and the second number is = 𝑥 + 15
according to the question –
𝑥 + 𝑥 + 15 = 95
2𝑥 + 15 = 95
2𝑥 = 95 – 15 (by transposing)
2𝑥 = 80
𝑥 = \(\displaystyle \frac{{80}}{2}\)
𝑥 = 40
so the first number 𝑥 = 40 answer
and the second number is = 𝑥 + 15
= 40 + 15
= 55 answer

5. Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Sol.: Let’s first number is = 5𝑥
and the second number is = 3𝑥
according to the question –
5𝑥- 3𝑥 = 18
2𝑥 = 18
𝑥 = \(\displaystyle \frac{{18}}{2}\)
𝑥 = 9
so the first number is 5𝑥 = 5 × 9
= 45 answer
and second number is 3𝑥 = 3 × 9
= 27 answer

6. Three consecutive integers add up to 51. What are these integers?
Sol.: Let’s first integers is = 𝑥
second integers is = 𝑥 + 1 (difference of two consecutive integers)
so third integers is = 𝑥 + 2
according to the question –
𝑥 + 𝑥 + 1 + 𝑥 + 2 = 51
3𝑥 + 3 = 51
3𝑥 = 51 – 3 (by transposing)
3𝑥 = 48
𝑥 = \(\displaystyle \frac{{48}}{3}\)
𝑥 = 16
so the first integers 𝑥 = 16 answer
second integers is 𝑥 + 1 = 16 + 1 = 17 answer
third integers is 𝑥 + 2 = 16 + 2 = 18 answer
These three consecutive integers are 16, 17 and 18.

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.
Sol.: Let’s first multiples of 8 is = 𝑥
second multiple of 8 is = 𝑥 + 8 (difference of two consecutive multiples of 8)
and third multiple of 8 is = 𝑥 + 16
according to the question –
𝑥 + 𝑥 + 8 + 𝑥 + 16 = 888
3𝑥 + 24 = 888
3𝑥 = 888 – 24
3𝑥 = 864
𝑥 = \(\displaystyle \frac{{888}}{3}\)
𝑥 = 288
So the first multiple of 8 is 𝑥 = 288
second multiple of 8 is 𝑥 + 8 = 288 + 8 = 296
third multiple of 8 is 𝑥 + 16 = 288 + 16 = 304
The three consecutive multiples of 8 is 288, 296 and 304.

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Sol.: Let’s first integers is = 𝑥
Second integers is = 𝑥 + 1
and third integers is = 𝑥 + 2
According to the question –
Respectively multiplied by 2, 3 and 4.
2 × 𝑥 + 3 × (𝑥 + 1) + 4 × (𝑥 + 2) = 74
2𝑥 + 3𝑥 + 3 + 4𝑥 + 8 = 74
9𝑥 + 11 = 74
9𝑥 = 74 – 11 (by transposing)
9𝑥 = 63
𝑥 = \(\displaystyle \frac{{63}}{9}\)
𝑥 = 7
So first integers is 𝑥 = 7
Second integers is 𝑥 + 1 = 7 + 1 = 8
and third integers is 𝑥 + 2 =  7 + 2 = 9
These three consecutive integers are 7. 8 and 9.

9. The ages of Rahul and Haroon are in the ratio 5 : 7. Four years later the sum of their ages will be 56 years. What are their present ages?
Sol.: Let’s Rahul age is = 5𝑥
and Haroon age is = 7𝑥
After 4 years Rahul age = 5𝑥 + 4
and Haroon age = 7𝑥 + 4
according to the question –
5𝑥 + 4 + 7𝑥 + 4 = 56
12𝑥 + 8 = 56
12𝑥 = 56 – 8 (by transposing)
12𝑥 = 48
𝑥 = \(\displaystyle \frac{{48}}{12}\)
𝑥 = 4
So, Rahul age is 5𝑥 = 5 × 4 = 20 year
and Haroon age is 7𝑥 = 7 × 4 = 28 year 

10. The number of boys and girls in a class are in the ratio 7 : 5. The number of boys is 8 more than the number of girls. What is the total class strength?
Sol.: Let’s the number of girls are = 𝑥
and the number of boys are = 𝑥 + 8
according to the question –
\(\displaystyle \frac{{x+8}}{x}=\frac{7}{5}\)
\(\displaystyle 7\times x=5\times (x+8)\) (by cross multiplication)
\(\displaystyle \begin{array}{l}7x=5x+40\\7x-5x=40\\2x=40\end{array}\)
𝑥 = \(\displaystyle \frac{{40}}{2}\)
𝑥 = 20
So the number of girls are 𝑥 = 20
and the number of boys are 𝑥 + 8 = 20 + 8 = 28
the total class strength is 20 + 28 = 48 answer

11. Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Sol.: Let’s Baichung father’s age is = 𝑥
Baichung grandfather’s age = 𝑥 + 26
and Baichung’s age = 𝑥 – 29
according to the question –
𝑥 + 𝑥 + 26 + 𝑥 – 29 = 135
3𝑥 – 3 = 135
3𝑥 = 135 + 3 (by transposing)
3𝑥 = 138
𝑥 = \(\displaystyle \frac{{138}}{3}\)
𝑥 = 46
So, Baichung’s father age 𝑥 = 46 years
Baichung’s grandfather age 𝑥 + 26
= 46 + 26 = 72 years
Baichung’s age 𝑥 – 29
= 46 – 29 = 17 years

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Sol.: Let’s Ravi’s present age is = 𝑥
after 15 years Ravi’s age = 𝑥 + 15
according to the question –
𝑥 + 15 = 4𝑥
𝑥 – 4x = – 15 (by transposing)
– 3𝑥 = – 15
𝑥 = \(\displaystyle \frac{{-15}}{-3}\)
𝑥 = 5
So Ravi’s present age is = 5 years

13. A rational number is such that when you multiply it by \(\displaystyle \frac{5}{2}\) and add \(\displaystyle \frac{2}{3}\) to the product, you get \(\displaystyle -\frac{7}{{12}}\). What is the number?
Sol.: Let’s the rational number is = 𝑥
x is multiply by \(\displaystyle \frac{5}{2}\) then product is
\(\displaystyle x\times \frac{5}{2}\)
\(\displaystyle \frac{5x}{2}\)
adding \(\displaystyle \frac{2}{3}\) then
according to the question –
\(\displaystyle \frac{{5x}}{2}+\frac{2}{3}=-\frac{7}{{12}}\)
\(\displaystyle \frac{{15x+4}}{6}=-\frac{7}{{12}}\) (by L.C.M)
\(\displaystyle 12\times (15x+4)=-7\times 6\) (by cross multiplication)
\(\displaystyle \begin{array}{l}180x+48=-42\\180x=-42-48\\180x=-90\\x=\frac{{-90}}{{180}}\end{array}\)
\(\displaystyle x=\frac{{-1}}{2}\)
So rational number is \(\displaystyle \frac{1}{2}\) answer

14. Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2 : 3 : 5. The total cash with Lakshmi is 4,00,000. How many notes of each denomination does she have?
Sol.: ∵ Ratio of ₹ 100 notes is = 2
Let’s number of ₹ 100 notes = 2𝑥
So, Value of ₹ 100 notes = 2𝑥 × 100 = 200𝑥 
∵ Ratio of ₹ 50 notes is = 3
Let’s number of ₹ 50 notes = 3𝑥
So, Value of ₹ 50 notes = 3𝑥 × 50 = 150𝑥
∵ Ratio of ₹ 10 notes is = 5
Let’s number of ₹ 10 notes = 5𝑥
So, Value of ₹ 10 notes = 5𝑥 × 10 = 50𝑥 
according to the question – 
200𝑥 + 150𝑥 + 50𝑥 = 400000
400𝑥 = 400000
\(\displaystyle x=\frac{{400000}}{{400}}\)
𝑥 = 1000
So the number of ₹100 notes = 2𝑥 = 2 × 1000 = 2000 notes
number of ₹ 50 notes = 3𝑥 = 3 × 1000 = 3000 notes
number of ₹ 10 notes = 5𝑥 = 5 × 1000 = 5000 notes

15. I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Sol.: Let’s number of coins ₹ 5 is = x
and number of coins ₹ 2 is = 3x
Total number of coins is = 160
Then number of coins ₹ 1 is = (160 – 3x – x) = (160-4x)
Value of ₹ 1 coins = (160 – 4x) × 1 = 160 – 4x
Value of ₹ 2 coins = 3x × 2 = 6x
Value of ₹ 5 coins = x × 5 = 5x
According to the question –
160 – 4x + 6x + 5x = 300
160 – 4x + 11x = 300
7x = 300 – 160 (by transposing)
7x = 140
\(\displaystyle x=\frac{{140}}{{7}}\)
x = 20
So, Number of ₹ 5 coins is = x
= 20 coins
Number of ₹ 2 coins is = 3x
= 3 × 20
= 60 coins
Number of ₹ 1 coins is = 160 – 4x
= 160 – 4 × 20
= 160 – 80 
= 80 coins

16. The organisers of an essay competition decide that a winner in the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Sol.: Total number of participants is = 63
Let’s number of winner is = x
and number of participant who does not win = 63 – x
Each winner gets a prize of = ₹ 100
So total winner prize money = x × 100
= 100x
Same, each participant who does not win gets a prize of = ₹ 25
So total each participant who does not win gets a prize money = (63 – x) × 25
= 1575 – 25x
According to the question –
100x + 1575 – 25x = 3000
75x = 3000 – 1575 (by transposing)
75x = 1425
\(\displaystyle x=\frac{{1575}}{{75}}\)
x = 19
So the number of winners is 19 answer.

Ncert Solutions for Class 8 Maths Chapter 2 EXERCISE 2.3

Solve the following equations and check your results.

1. 3x = 2x + 18 2. 5t – 3 = 3t – 5 3. 5x + 9 = 5 + 3x
4. 4z + 3 = 6 + 2z 5. 2x – 1 = 14 – x 6. 8x + 4 = 3 (x – 1) + 7
7. \(\displaystyle x=\frac{4}{5}(x+10)\) 8. \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\) 9. \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)
10. \(\displaystyle 3m=5m-\frac{8}{5}\)    

Solutions :

1. 3x = 2x + 18
Sol.: 3x = 2x + 18
3x – 2x = 18 (by transposing)
x = 18 answer 

2. 5t – 3 = 3t – 5
Sol.: 5t – 3 = 3t – 5
5t – 3t = – 5 + 3
2t = – 2
\(\displaystyle t=-\frac{2}{2}\)
t = – 1 answer 

3. 5x + 9 = 5 + 3x
Sol.: 5x + 9 = 5 + 3x
5x – 3x = 5 – 9 (by transposing)
2x = – 4
\(\displaystyle x=-\frac{4}{2}\)
x = – 2 answer 

4. 4z + 3 = 6 + 2z
Sol.: 4z + 3 = 6 + 2z
4z – 2z = 6 – 3 (by transposing)
2z = 3
\(\displaystyle z=\frac{{3}}{{2}}\) answer

5. 2x – 1 = 14 – x
Sol.: 2x – 1 = 14 – x
2x + x = 14 + 1 (by transposing)
3x = 15
\(\displaystyle x=\frac{{15}}{{3}}\)
x = 5

6. 8x + 4 = 3 (x – 1) + 7
Sol.: 8x + 4 = 3 (x – 1) + 7
8x + 4 = 3x – 3 + 7
8x – 3x = – 3 + 7 –  4 (by transposing)
5x = 7 – 7
5x = 0
\(\displaystyle x=\frac{{0}}{{5}}\)
x = 0 answer

7. \(\displaystyle x=\frac{4}{5}(x+10)\)
Sol.: \(\displaystyle x=\frac{4}{5}(x+10)\)
\(\displaystyle \begin{array}{l}x=\frac{{4x}}{5}+\frac{4}{5}\times 10\\x=\frac{{4x}}{5}+\frac{8}{1}\end{array}\)
\(\displaystyle x=\frac{{4x+40}}{5}\) (by L.C.M.)
\(\displaystyle 5x=4x+40\)
\(\displaystyle 5x-4x=40\) (by transposing)
x = 40 answer

8. \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\)
Sol.: \(\displaystyle \frac{{2x}}{3}+1=\frac{{7x}}{{15}}+3\)
\(\displaystyle \frac{{2x}}{3}-\frac{{7x}}{{15}}=3-1\) (by transposing)
\(\displaystyle \frac{{10x-7x}}{{15}}=\frac{2}{1}\) (by cross multiplication)
10x – 7x = 2 × 15
3x = 30
\(\displaystyle x=\frac{{30}}{3}\)
x = 10 answer

9. \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)
Sol.: \(\displaystyle 2y+\frac{5}{3}=\frac{{26}}{3}-y\)
\(\displaystyle 2y+y=\frac{{26}}{3}-\frac{5}{3}\) 
\(\displaystyle \begin{array}{l}3y=\frac{{26-5}}{3}\\3y=\frac{{21}}{3}\end{array}\) (by transposing)
\(\displaystyle y=\frac{7}{3}\) answer

10. \(\displaystyle 3m=5m-\frac{8}{5}\)
Sol.: \(\displaystyle 3m=5m-\frac{8}{5}\)
\(\displaystyle 3m-5m=-\frac{8}{5}\) (by transposing)
\(\displaystyle -2m=-\frac{8}{5}\) (both side have – sign so – to – cancel)
\(\displaystyle m=\frac{8}{{2\times 5}}\)
\(\displaystyle m=\frac{4}{5}\) answer

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